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I am looking for a reference that has a proof that a path groupoid is a groupoid internal to the category of diffeological spaces. I do know how to prove this fact, and a proof is not hard. My reason for asking this question is that I don't want to clutter a paper with things that are apparently well-known. However the nLab article linked to above gives no references.

The places I tried to look this fact up are: Diffeology by Patrick Iglesias-Zemmour, papers of Schreiber and Waldorf, papers of Picken with various collaborators, the nLab, google and google scholar.

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Might be folklore. As a side comment, there's a slightly better way to construct something equivalent, where you don't modify the paths (i.e. use sitting instants) but modify the composition: instead of using the standard diffeo [0,1] -> [0,2], use something that is flat at 1/2. The result seems to me so obvious that one doesn't really need a proof, just a comment. But that doesn't take care of attribution... –  David Roberts Aug 10 '13 at 0:24
David, I also suspect the result is folklore. I just wanted to make sure that it is. –  Eugene Lerman Aug 11 '13 at 16:43
David, I agree that a proof that the concatenation of paths is smooth is easy. The only possible subtlety is that in the groupoid you concatenate thin homotopy classes, and the space of classes has quotient diffeology. So at some point you need to check that in diffeological spaces fiber products commute with quotients, which they do. –  Eugene Lerman Aug 11 '13 at 16:46

1 Answer 1

If you mean the "Poincaré groupoid" made by fixed-end homotopy classes of smooth paths of a diffeological space, this is a honest diffeological groupoid. Its construction is contained in Chapter V of the book "Diffeology", exactly the paragraph titled "Poincaré Groupoid and Homotopy Groups" (beginning p.112). If you have something else in mind, be more explicit please.

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Thank you for the answer. No, the path groupoid I mean is the groupoid of equivalence classes of paths in a manifold related by the so called "thin" homotopies: the rank of the homotopy thought of as a smooth map from a square is required to be at most 1 at every point. To put it differently the homotopy sweeps no area. I thought I was being explicit by putting a link to a definition in ncatlab. I did not see this definition in your book. –  Eugene Lerman Dec 9 '13 at 20:48
@Eugene OK, I see. I wasn't aware of such a concept. I didn't get the definition by clicking on the link first time, it's why I had no idea on what you were referring to. I'll take a look. –  Patrick I-Z Dec 12 '13 at 3:45

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